Metamath Proof Explorer


Theorem seqomlem3

Description: Lemma for seqom . (Contributed by Stefan O'Rear, 1-Nov-2014)

Ref Expression
Hypothesis seqomlem.a 𝑄 = rec ( ( 𝑖 ∈ ω , 𝑣 ∈ V ↦ ⟨ suc 𝑖 , ( 𝑖 𝐹 𝑣 ) ⟩ ) , ⟨ ∅ , ( I ‘ 𝐼 ) ⟩ )
Assertion seqomlem3 ( ( 𝑄 “ ω ) ‘ ∅ ) = ( I ‘ 𝐼 )

Proof

Step Hyp Ref Expression
1 seqomlem.a 𝑄 = rec ( ( 𝑖 ∈ ω , 𝑣 ∈ V ↦ ⟨ suc 𝑖 , ( 𝑖 𝐹 𝑣 ) ⟩ ) , ⟨ ∅ , ( I ‘ 𝐼 ) ⟩ )
2 peano1 ∅ ∈ ω
3 fvres ( ∅ ∈ ω → ( ( 𝑄 ↾ ω ) ‘ ∅ ) = ( 𝑄 ‘ ∅ ) )
4 2 3 ax-mp ( ( 𝑄 ↾ ω ) ‘ ∅ ) = ( 𝑄 ‘ ∅ )
5 1 fveq1i ( 𝑄 ‘ ∅ ) = ( rec ( ( 𝑖 ∈ ω , 𝑣 ∈ V ↦ ⟨ suc 𝑖 , ( 𝑖 𝐹 𝑣 ) ⟩ ) , ⟨ ∅ , ( I ‘ 𝐼 ) ⟩ ) ‘ ∅ )
6 opex ⟨ ∅ , ( I ‘ 𝐼 ) ⟩ ∈ V
7 6 rdg0 ( rec ( ( 𝑖 ∈ ω , 𝑣 ∈ V ↦ ⟨ suc 𝑖 , ( 𝑖 𝐹 𝑣 ) ⟩ ) , ⟨ ∅ , ( I ‘ 𝐼 ) ⟩ ) ‘ ∅ ) = ⟨ ∅ , ( I ‘ 𝐼 ) ⟩
8 4 5 7 3eqtri ( ( 𝑄 ↾ ω ) ‘ ∅ ) = ⟨ ∅ , ( I ‘ 𝐼 ) ⟩
9 frfnom ( rec ( ( 𝑖 ∈ ω , 𝑣 ∈ V ↦ ⟨ suc 𝑖 , ( 𝑖 𝐹 𝑣 ) ⟩ ) , ⟨ ∅ , ( I ‘ 𝐼 ) ⟩ ) ↾ ω ) Fn ω
10 1 reseq1i ( 𝑄 ↾ ω ) = ( rec ( ( 𝑖 ∈ ω , 𝑣 ∈ V ↦ ⟨ suc 𝑖 , ( 𝑖 𝐹 𝑣 ) ⟩ ) , ⟨ ∅ , ( I ‘ 𝐼 ) ⟩ ) ↾ ω )
11 10 fneq1i ( ( 𝑄 ↾ ω ) Fn ω ↔ ( rec ( ( 𝑖 ∈ ω , 𝑣 ∈ V ↦ ⟨ suc 𝑖 , ( 𝑖 𝐹 𝑣 ) ⟩ ) , ⟨ ∅ , ( I ‘ 𝐼 ) ⟩ ) ↾ ω ) Fn ω )
12 9 11 mpbir ( 𝑄 ↾ ω ) Fn ω
13 fnfvelrn ( ( ( 𝑄 ↾ ω ) Fn ω ∧ ∅ ∈ ω ) → ( ( 𝑄 ↾ ω ) ‘ ∅ ) ∈ ran ( 𝑄 ↾ ω ) )
14 12 2 13 mp2an ( ( 𝑄 ↾ ω ) ‘ ∅ ) ∈ ran ( 𝑄 ↾ ω )
15 8 14 eqeltrri ⟨ ∅ , ( I ‘ 𝐼 ) ⟩ ∈ ran ( 𝑄 ↾ ω )
16 df-ima ( 𝑄 “ ω ) = ran ( 𝑄 ↾ ω )
17 15 16 eleqtrri ⟨ ∅ , ( I ‘ 𝐼 ) ⟩ ∈ ( 𝑄 “ ω )
18 df-br ( ∅ ( 𝑄 “ ω ) ( I ‘ 𝐼 ) ↔ ⟨ ∅ , ( I ‘ 𝐼 ) ⟩ ∈ ( 𝑄 “ ω ) )
19 17 18 mpbir ∅ ( 𝑄 “ ω ) ( I ‘ 𝐼 )
20 1 seqomlem2 ( 𝑄 “ ω ) Fn ω
21 fnbrfvb ( ( ( 𝑄 “ ω ) Fn ω ∧ ∅ ∈ ω ) → ( ( ( 𝑄 “ ω ) ‘ ∅ ) = ( I ‘ 𝐼 ) ↔ ∅ ( 𝑄 “ ω ) ( I ‘ 𝐼 ) ) )
22 20 2 21 mp2an ( ( ( 𝑄 “ ω ) ‘ ∅ ) = ( I ‘ 𝐼 ) ↔ ∅ ( 𝑄 “ ω ) ( I ‘ 𝐼 ) )
23 19 22 mpbir ( ( 𝑄 “ ω ) ‘ ∅ ) = ( I ‘ 𝐼 )